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**Unformatted text preview: **SOLUTIONS TO MATH 51 MIDTERM 1 January 29, 2004 1. Find all solutions of the following system: x 1- x 2 + x 3 + 2 x 4 = 3 x 2 + x 3 + x 4 = 3 x 1 + x 2 + 3 x 3 + 4 x 4 = 9 Solution. Write the augmented matrix and then use Gaussian elimination: 1- 1 1 2 3 1 1 1 3 1 1 3 4 9 1- 1 1 2 3 1 1 1 3 2 2 2 6 1 0 2 3 6 0 1 1 1 3 0 0 0 0 0 So x 1 + 2 x 3 + 3 x 4 = 6 x 2 + x 3 + x 4 = 3 The free variables are x 3 and x 4 , so the solutions are: x 1 = 6- 2 x 3- 3 x 4 , x 2 = 3- x 3- x 4 , ( x 3 R , x 4 R ) 2. Let L be the intersection of the two planes x + y + z = 4 and 2 x + 3 y + z = 9 . Find a parametric equation for L . Solution. Write the augmented matrix and use Gaussian elimination: fl fl fl fl 1 1 1 4 2 3 1 9 fl fl fl fl fl fl fl fl 1 1 1 4 0 1- 1 1 fl fl fl fl fl fl fl fl 1 0 2 3 0 1- 1 1 fl fl fl fl . so x + 2 z = 3 and y- z = 1, or (moving the free variable z to the right hand side) x = 3- 2 z and y = 1 + z . Thus the intersection is given by x y z = 3- 2 z 1 + z z = 3 1 + z - 2 1 1 . 3(a) . Suppose u , v , and w are points in R n such that k u k = k v k = k w k = 1 and such that w =- u . Suppose also that v is not equal to u or to w . Prove that the triangle uvw has a right angle at v . 1 Solution. The vector from u to v is v- u . The vector from w to v is v- w . We want to show that these two vectors are orthogonal, so we calculate the dot product: ( v- u ) ( v- w ) = ( v- u ) ( v + u ) = v v + v u- u v- u u = k v k 2-k u k 2 = 1 2- 1 2 = 0 3(b) . Suppose x , y , and z are vectors in R n whose norms are 1, 2, and 3, respec- tively. Suppose each vector is orthogonal (i.e., perpendicular) to each of the other two. Find a scalar c such that the vector x + c y- z is orthogonal to the vector x + y + z . Solution The vectors x + y + z and x + c y- z vectors will be orthogonal provided their dot product is 0. When we multiply it out (i.e., use the distributive property), all the mixed terms ( x y , x z , etc.) are 0 by orthogonality. So 0 = ( x + y + z ) (...

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